Answer
$$\frac{x}{{5\sqrt {{x^2} + 5} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{{\left( {{x^2} + 5} \right)}^{3/2}}}}} dx \cr
& {\text{Refer to the triangle below}} \cr
& \tan \theta = \frac{x}{{\sqrt 5 }},{\text{ }}x = \sqrt 5 \tan \theta ,{\text{ }}dx = \sqrt 5 {\sec ^2}\theta d\theta \cr
& \int {\frac{1}{{{{\left( {{x^2} + 5} \right)}^{3/2}}}}} dx = \int {\frac{{\sqrt 5 {{\sec }^2}\theta d\theta }}{{{{\left( {{{\left( {\sqrt 5 \tan \theta } \right)}^2} + 5} \right)}^{3/2}}}}} \cr
& = \int {\frac{{\sqrt 5 {{\sec }^2}\theta }}{{{{\left( {5{{\tan }^2}\theta + 5} \right)}^{3/2}}}}} d\theta \cr
& = \int {\frac{{\sqrt 5 {{\sec }^2}\theta }}{{{5^{3/2}}{{\left( {{{\tan }^2}\theta + 1} \right)}^{3/2}}}}} d\theta \cr
& = \int {\frac{{\sqrt 5 {{\sec }^2}\theta }}{{{5^{3/2}}{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}} d\theta \cr
& = \frac{1}{5}\int {\frac{1}{{\sec \theta }}} d\theta \cr
& = \frac{1}{5}\int {\cos \theta } d\theta \cr
& {\text{Integrate, }} \cr
& = \frac{1}{5}\sin \theta + C \cr
& {\text{Refer to the triangle, }}\sin \theta = \frac{x}{{\sqrt {{x^2} + 5} }} \cr
& = \frac{1}{5}\left( {\frac{x}{{\sqrt {{x^2} + 5} }}} \right) + C \cr
& = \frac{x}{{5\sqrt {{x^2} + 5} }} + C \cr} $$
